Question 1047218
{{{(1/2)*(1/2)*(1/2)*(1/2) = 1/16}}}
is the probability that they will have
girl, girl, girl, girl, in that order.
It is also the probability of
boy, boy, boy, boy, in that order,
and it is the probability of any sequence of 4 children,
such as
boy, boy, girl, girl, or
girl, girl, boy, boy, or
boy, girl, boy, girl, or
girl, boy, girl, boy, or
boy, girl, girl, boy, or
girl, boy, boy,girl.
So out of the {{{16}}} sequences possible
there are {{{6}}} sequences that would give them exactly 2 girls and 2 boys.
The probability of that is {{{6/16=highlight(3/8=0.375="37.5%")}}} .
When you are faced with problems of that sort,
think of {{{(boy+girl)^4}}} or {{{(a+b)^4}}} .

In the painful {{{(a+b)^4=(a+b)*(a+b)*(a+b)*(a+b)}}} multiplication,
you would end with {{{2*2*2*2=2^4=16}}} products,
including {{{red(4)}}} {{{a^3b}}} product terms, {{{green(6)}}} {{{a^2b^2}}} , and {{{blue(4)}}} {{{ab^3}}} ,
but then you "collect like terms" to get
{{{(a+b)^4=a^4+red(4)a^3b+green(6)a^2b^2+blue(4)ab^3+b^4}}} .
The term {{{green(6)a^2b^2}}} tells you that there were {{{green(6)}}}
outcomes where 2 a's and 2 b's happened, in various orders.
The probability of that happening is {{{green(6)/16}}} .